Answer:
the inflection points are

So,

It is concave down at the intervals
![(-\infty , -0.85] \cup [-0.14,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%20%2C%20-0.85%5D%20%5Ccup%20%5B-0.14%2C%5Cinfty%29)
And it would be concave up at

Step-by-step explanation:
Remember that to find inflection points you need to find where

Since

Then using the product and the chain rule you have that

And then, using again the chain rule and the product rule you have that

Therefore you have to solve the equation

Using the quadratic equation you get that there are two solutions, so the inflection points are

So,

Now remember that a function is concave up if the derivative is greater than zero and concave down if the derivative is less than zero. Therefor you have to solve these inequalties

And you would get that is concave down at the intervals
![(-\infty , -0.85] \cup [-0.14,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%20%2C%20-0.85%5D%20%5Ccup%20%5B-0.14%2C%5Cinfty%29)
And it would be concave up at
I believe it is Row operations I took that class last year
Answer:
(-∞,7) U (7,∞)
Step-by-step explanation:
f(x)= x+2
g(x) = x-7

Here we have x-7 in the denominator
To find domain we set the denominator =0 and solve for x
x-7=0
Add 7 on both sides
x=7
x=7 makes the denominator 0 that is undefined
So we ignore 7 for x
Hence domain is
(-∞,7) U (7,∞)
Answer:
11
Step-by-step explanation:
4^3/2 - 7 × 3
64/2 - 21
32 - 21 = 11